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Question 47 of 50

Q.(i) Calculate the oxidation number of central metal atom of the following complex compound: [Cr(NH3)5Cl]SO4.

(ii) When aqueous solution of one mole of CrCl3.6H2O complex compound is treated with excess of AgNO3, 3 moles of AgCl are precipitated. Determine the formula of the complex compound. [1+1]
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2024Subjective· 2mImportance★★★★★
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Balancing charges gives Cr = +3 in the ammine-chloro complex; all-Cl-precipitated-by-AgNO3_3 means the formula is the hexaaqua complex with 3 ionisable chlorides.

(i) Oxidation number of Cr in [Cr(NH3_3)5_5Cl]SO4_4:

NH3_3 is a neutral ligand (charge 0), Cl−^- within the coordination sphere carries −1, and SO42−_4^{2-} is the counter-ion outside the sphere, balancing a +2 charge on the complex ion [Cr(NH3_3)5_5Cl]2+^{2+}.

Let oxidation number of Cr = xx:

x+5(0)+(−1)=+2  ⇒  x=+3x + 5(0) + (-1) = +2 \;\Rightarrow\; x = +3

(ii) Formula from the AgNO3_3 precipitation: …

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